I am trying to compute mean and variance of the stochastic process $X_t$, which is a Brownian bridge from x to y, in the time-interval $[t,T]$.
$$X_t = y + (x-y)(\frac{T-t}{T})^k+\sigma\int_0^t(\frac{T-t}{T-s})^kdW_s$$
For the mean I try to argue like this: $$E[X_t] = y+ (x-y)(\frac{T-t}{T})^k+E[\sigma\int_0^t(\frac{T-t}{T-s})^kdW_s]=y+ (x-y)(\frac{T-t}{T})^k$$ since $E[\sigma\int_0^t(\frac{T-t}{T-s})^kdW_s]\leq E[\sigma\int_0^t(\frac{T-t}{T-t})^kdW_s]=E[\sigma\int_0^tdW_s]=0$, considering $s\leq t \leq T$.
Computing the variance, I am stuck the following way: $$E[X_t^2]=E[( y + (x-y)(\frac{T-t}{T})^k+\sigma\int_0^t(\frac{T-t}{T-s})^kdW_s)^2]$$ multiplying out $(a+c+b)^2$ will create a lot of terms: $$E[yy + y(x-y)(\frac{T-t}{T})^k+y\sigma\int_0^t(\frac{T-t}{T-s})^kdW_s + \\ y(x-y)(\frac{T-t}{T})^k + (x-y)^2(\frac{T-t}{T})^{2k}+(x-y)(\frac{T-t}{T})^k\sigma\int_0^t(\frac{T-t}{T-s})^kdW_s] + \\ ... $$
Is there a direct way to solve this?
Note that the Itô integral is not monotone, i.e. $f \leq g$ does not imply $$\int_0^t f(s) \, dW_s \leq \int_0^t g(s) \, dW_s$$ and therefore the estimate used for the calculation of the expectation does not work out. Instead, use that
$$(t,\omega) \mapsto M_t(\omega) := \left(\int_0^t f(s) \, dB_s \right)(\omega)$$
is a martingale for any bounded (progressively measurable) function $f$. In particular $\mathbb{E}M_t = \mathbb{E}M_0$.
Concerning the variance: By definition,
$$\text{var} \, X_t = \mathbb{E}((X_t-\mathbb{E}X_t)^2) = \sigma^2 \mathbb{E} \left[\left(\int_0^t \, \left( \frac{T-t}{t-s} \right)^k \, dW_s \right)^2 \right]$$
so since you substract the mean, most of the terms vanish. Now apply Itô's isometry.